S2T2

Jim Randell 11:16 am on 21 June 2022 Permalink Reply
Tags: by: Eric Emmet ( 10 )   

• 

  Jim Randell 11:17 am on 21 June 2022 Permalink | Reply

  This is similar to Puzzle 15 (and Puzzle 66).

  Some obvious initial analysis:

  Statement B1 (“The wages of one of us is different from the sum of those of the other two”) must always be true, as it it not possible for each of the three to have wages that are equal to the sum of the other two. (Who would have the largest value?)

  So B cannot be a Wotta-Woppa. (And so A1 is false).

  Which means B3 must be also be true. So C must make more true statements than B, so B cannot be a Pukka. Hence B is a Shilli-Shalla, and C is a Pukka. (And B2 is false).

  And so C3 must be true, and “Stupid” is a Shilli-Shalla.

  And as A1 is false, then A cannot be a Pukka, and A3 must also be false.

  I incorporated these findings into a generic solution to get a program that runs in 71ms.

  Run: [ @replit ]

  from enigma import (subsets, tuples, irange, product, div, intersect, printf)
  
  # Pukka - always tell the truth
  def P(ss):
    return all(ss)
  
  # Wotta-Woppa - never tells the truth
  def W(ss):
    return not any(ss)
  
  # Shilli-Shalla - alternates between true and false
  def S(ss):
    return all(a != b for (a, b) in tuples(ss, 2))
  
  # wages, by tribe
  wage = { P: irange(10, 19), S: irange(20, 29), W: irange(30, 39) }
  
  # assign tribes to the participants
  for ts in ((W, S, P), (S, S, P)):
    (A, B, C) = ts
  
    # B's statements
    Bs = [True, False, True]
    assert B(Bs)
  
    # assign names to the participants
    for ns in subsets("STU", size=3, select="P"):
      (nA, nB, nC) = ns
      if nC == 'S': continue  # not possible
      (iS, iT, iU) = (ns.index(x) for x in "STU")
  
      # assign wages to the participants
      for ws in product(wage[A], wage[B], wage[C]):
        (wA, wB, wC) = ws
  
        # check A's statements
        As = [False, div(10 * wA, 9) in {wB, wC}, wA == wB + 10]
        if not A(As): continue
  
        # check C's statements
        Cs = [
          ws[iU] > ws[iT],
          ts[iS] is S,
          bool(intersect([{ div(85 * x, 100) for x in ws }, ws])),
        ]
        if not C(Cs): continue
  
        # check B's 3rd statement
        assert Cs.count(True) > Bs.count(True)
  
        # output solution
        f = lambda x: x.__name__
        printf("A={A} {nA} {wA}; B={B} {nB} {wB}; C={C} {nC} {wC} [{As}; {Bs}; {Cs}]", A=f(A), B=f(B), C=f(C))
  

  Solution: C is Toothless, he is a Pukka, his wage is 17 Hopes.

  A is Ugly, he is a Wotta-Woppa, his wage is 31 – 39 Hopes.

  B is Stupid, he is a Shilli-Shalla, his wage is 20 Hopes.

  C‘s wage (17) is 15% less than that of B (20).

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  • 

    Frits 3:04 pm on 21 June 2022 Permalink | Reply

    @Jim, You swapped the ranges of Wotta-Woppa and Shilli-Shalla.

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    • 

      Jim Randell 3:10 pm on 21 June 2022 Permalink | Reply

      Thanks. Fixed now.

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  • 

    Jim Randell 12:25 pm on 27 June 2022 Permalink | Reply

    Inspired by Frits’ solution (below) I wrote a solution that uses the [[ SubstitutedExpression ]] solver from the enigma.py library, and is just a restatement of the puzzle text with no analysis.

    The values assigned for the tribes make it easy to express the wages alphametically.

    This Python program runs in 73ms.

    Run: [ @replit ]

    from enigma import (SubstitutedExpression, irange, tuples, div, printf)
    
    #    statements  tribe  name  wage
    # A: a d g       p      x     ps
    # B: b e h       q      y     qt
    # C: c f i       r      z     ru
    #
    # statements: 0 (false), 1 (true)
    # tribe: 1 (P), 2 (SS), 3 (WW)
    # name: 1 (S), 2 (T), 3 (U)
    # wage (units): 0 - 9
    
    # check statements for a tribe
    def tribe(t, ss):
      if t == 1: return all(ss)
      if t == 2: return all(a != b for (a, b) in tuples(ss, 2))
      if t == 3: return not any(ss)
    
    # B1: one of the wages is different from the sum of the other two
    def B1(*ws):
      t = sum(ws)
      return any(w != t - w for w in ws)
    
    # C1: Ugly (3) earns more than Toothless (2)
    def C1(ns, ws):
      d = dict(zip(ns, ws))
      return d[3] > d[2]
    
    # C2: the wages of one of us is 85% the wages of another
    def C2(*ws):
      for w in ws:
        x = div(85 * w, 100)
        if x in ws: return True
      return False
    
    # expressions to evaluate
    exprs = [
      # tribes and statements
      "tribe({p}, [{a}, {d}, {g}])",
      "tribe({q}, [{b}, {e}, {h}])",
      "tribe({r}, [{c}, {f}, {i}])",
    
      # A1: "B is a Wotta-Woppa (3)" = {a}
      "int({q} == 3) = {a}",
      # A2: "My wages are [0.9] one of the others" = {d}
      "int(div({ps} * 10, 9) in { {qt}, {ru} }) = {d}",
      # A3: "I get 10 Hopes more than B" = {g}
      "int({ps} == {qt} + 10) = {g}",
      # B1: "The wages of one of us is different from the sum of the wages
      # of the other two" = {b}
      "int(B1({ps}, {qt}, {ru})) = {b}",
      # B2: "We all belong the same tribe" = {e}
      "int({p} == {q} == {r}) = {e}",
      # B3: "More of C's statements are true than mine" = {h}
      "int({c} + {f} + {i} > {b} + {e} + {h}) = {h}",
      # C1: "Ugly earns more than Toothless" = {c}
      "int(C1([{x}, {y}, {z}], [{ps}, {qt}, {ru}])) = {c}",
      # C2: "The wages of one of us are 0.85 the wages of another" = {f}
      "int(C2({ps}, {qt}, {ru})) == {f}",
      # C3: "Stupid (1) is a Shilli-Shalla (2)" = {i}
      "int({ {x}: {p}, {y}: {q}, {z}: {r} }[1] == 2) = {i}",
    ]
    
    # invalid digit / symbol assignments
    d2i = dict()
    for d in irange(0, 9):
      vs = set()
      if d > 1: vs.update('abcdefghi')
      if d < 1 or d > 3: vs.update('pqrxyz')
      d2i[d] = vs
    
    # create the puzzle
    p = SubstitutedExpression(
      exprs,
      distinct="xyz",
      d2i=d2i,
      env=dict(tribe=tribe, B1=B1, C1=C1, C2=C2),
      verbose=0,
    )
    
    # solve the puzzle and output solutions
    Ts = { 1: "Pukka", 2: "Shilli-Shalla", 3: "Wotta-Woppa" }
    Ns = { 1: "Stupid", 2: "Toothless", 3: "Ugly" }
    for s in p.solve():
      (a, b, c, d, e, f, g, h, i, p, q, r, s, t, u, x, y, z) = (s[k] for k in 'abcdefghipqrstuxyz')
      printf("A: {N:9s}  {T:13s}  {p}{s}  [{a} {d} {g}]", N=Ns[x], T=Ts[p])
      printf("B: {N:9s}  {T:13s}  {q}{t}  [{b} {e} {h}]", N=Ns[y], T=Ts[q])
      printf("C: {N:9s}  {T:13s}  {r}{u}  [{c} {f} {i}]", N=Ns[z], T=Ts[r])
      printf()
    

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    • 

      Frits 1:31 pm on 29 June 2022 Permalink | Reply

      Nice idea to let tribes go from 1 to 3 so you can use the tribe value as a tens unit of the wage.

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• 

  Frits 3:30 pm on 21 June 2022 Permalink | Reply

  For all tribes the veracity of the first and third statement must be the same.

     
  from enigma import SubstitutedExpression
  
  # tribes 0 = Wotta-Woppa, 1 = Shilli-Shilla, 2 = Pukka 
  # names: 0 = Stupid, 1 = Toothless, 2 = Ugly
  
  #     stmnts  tribe  wages   name
  #      0/1    0/1/2         0/1/2
  # A:  D E D     P     UV      K
  # B:  F G F     Q     WX      L
  # C:  H I H     R     YZ      M
  # the alphametic puzzle
  p = SubstitutedExpression(
    [
      # if Wotta-Woppa all statements are false
      "P != 0 or D == E == 0",
      "Q != 0 or F == G == 0",
      "R != 0 or H == I == 0",
      
      # if Shilli-Shalla all statements are alternating true/false
      "P != 1 or D + E == 1",
      "Q != 1 or F + G == 1",
      "R != 1 or H + I == 1",
      
      # if Pukka all statements are true
      "P != 2 or D == E == 1",
      "Q != 2 or F == G == 1",
      "R != 2 or H == I == 1",
      
      # ranges: Pukka 10-19, Shilli-Shalla 20-29 and Wotta-Woppa 30-39
      "(P != 0 or U == 3) and (P != 1 or U == 2) and (P != 2 or U == 1)",
      "(Q != 0 or W == 3) and (Q != 1 or W == 2) and (Q != 2 or W == 1)",
      "(R != 0 or Y == 3) and (R != 1 or Y == 2) and (R != 2 or Y == 1)",
      
      # A1: B is a Wotta-Woppa
      "(F == G == 0) == D",
      # A2: My wages are 25% less than 20% more than one of the others
      "(UV in {0.9 * WX, 0.9 * YZ}) == E",
      # A3: I get 10 hopes more than B
      "(UV == WX + 10) == D",
      
      # B1: The wages of one of us is different from the sum of those 
      # of the other two
      "(UV != WX + YZ or UV + WX != YZ or UV + YZ != WX) == F",
      # B2: We all belong to the same tribe
      "(P == Q == R) == G",
      # B3: More of C's statements are true than mine
      "(2* H + I >  2 * F + G) == F",
      
      
      # C1: Ugly earns more than Toothless
      "((UV if K == 2 else WX if L == 2 else YZ) > \
        (UV if K == 1 else WX if L == 1 else YZ)) == H",
      # C2: The wages of one of us are 15% less than the wages of another
      "((20 * UV in {17 * YZ, 17 * WX}) or \
        (20 * WX in {17 * YZ, 17 * UV}) or \
        (20 * YZ in {17 * UV, 17 * WX})) == I",
      # C3: Stupid is a Shilli-Shalla.
      "((K == 0 and P == 1) or \
        (L == 0 and Q == 1) or \
        (M == 0 and R == 1)) == H",
    ],
    answer="(D, E, F, G, H, I), (P, Q, R), (UV, WX, YZ), (K, L, M)",
    d2i=dict([(0, "UWY")] +
             [(2, "DEFGHI")] +
             [(3, "DEFGHIKLMPQR")] +
             [(k, "DEFGHIKLMPQRUWY") for k in range(4, 10)]),
    distinct="KLM",
    verbose=0,
  )
  
  # print answers
  for (_, ans) in p.solve():
    print(f"{ans}")
  

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